Two reactions are given below:
2 Fe₍ₛ₎+3/2 O_(2( g)) → Fe₂ O₃₍ₛ₎, Δ H°=-822 kJ / mol ; C₍ₛ₎+1/2 O_(2( g)) → CO_((g)), Δ H°=-110 kJ / mol
Then enthalpy change for following reaction 3 C₍ₛ₎+Fe₂ O₃₍ₛ₎ → 2 Fe₍ₛ₎+3 CO_((g)) is _______ kJ / mol.

Solution

Chemistry ma’am · 4 steps

Turn the iron reaction around and use the carbon reaction three times; then add the two ΔH values.

1. Reverse equation 1. In the target, Fe₂O₃ is a reactant, so run the first reaction backwards. Reversing a reaction flips the sign of Δ H: Fe₂O₃ → 2Fe + 3/2 O₂, Δ H = +822 kJ mol⁻¹.
2. Triple equation 2. The target needs 3 CO, so multiply the second reaction by 3. Multiplying a reaction multiplies its Δ H: 3C + 3/2 O₂ → 3CO, Δ H = 3 × (-110) = -330 kJ mol⁻¹.
3. Add them. On adding, the 3/2 O₂ appears on both sides and cancels. What remains is exactly the target reaction, so its Δ H is the sum of the two.

   Δ H = (+822) + (-330) = +492 kJ mol⁻¹

4. So. The reduction of Fe₂O₃ by carbon is endothermic, with Δ H = 492 kJ mol⁻¹.

The common slip: 712. Using the carbon reaction only once instead of three times gives 822 - 110 = 712, not 492.

The idea to keep. In Hess's law, match each species to the target first. Reverse a reaction and its ΔH changes sign; multiply a reaction and its ΔH is multiplied by the same number.

Answer from the NTA final answer key. Solution by PrepWiser.